Question

The equation of tangent to the curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ at the point $\left(a,b\right)$ is

• A. $\frac{x}{a}=-\frac{y}{b}$
• B. $\frac{x}{a}+\frac{y}{b}=2$
• C. $\frac{x}{a}=\frac{y}{b}$
• D. $\frac{x}{a}+\frac{y}{b}=n$

Answer 1

Answer: (B)

We have,
${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$
On differentiating w.r.t. $X$, we get
$\frac{n{x}^{n-1}}{a^n}+\frac{n{y}^{n-1}}{b^n}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-b^n{x}^{n-1}}{a^n{y}^{n-1}}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(a,b)}=\frac{-b^n{a}^{n-1}}{a^n{b}^{n-1}}=\frac{-b}{a}$
Equations of tangent at $(a,b)$ is
$y-b=\frac{-b}{a}(x-a)$
$\Rightarrow ay-ab=-bx+ab$
$\Rightarrow bx+ay=2ab$
$\Rightarrow \frac{bx}{ab}+\frac{ay}{ab}=\frac{2ab}{ab}$
$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$